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A capacitor of 4 muF is connected as s...

A capacitor of 4 `mu`F is connected as shown in the circuit as per figure. The internal resistance of the batlery is 0.5 `Omega`. The amount of charge on the capacitor plates will be:


(a) `0 mu C `
(b) `4 mu C `
(c) `16 muC `
(d) `8 mu C `

A

`0 mu C `

B

`4 mu C `

C

`16 muC `

D

`8 mu C `

Text Solution

Verified by Experts

The correct Answer is:
C
No current will flow in capacitor due to D.C. battery. Hence, current flow from resistance of 2`Omega` is
`I = (V)/(R+r) = (2.5)/(2+0.5)=(2.5)/(2.5) =1 A `
The potential difference across 2 `Omega`
`V= IR = I xx2 = 2 V `
Capacitor is parallel to resistance 2`Omega` and voltage drop around resistance 10 `Omega` is zero. Hence, p.d. across capacitor would be 2 V.
The charge on capacitor
`q = CV = 4xx10^(-6) xx2 = 8 xx10^(-6)` C
`:. q = 8 mu C `
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